# What is the equation of the quadratic graph with a focus of (4, 3) and a directrix of y = 13?

**Solution:**

Let P(x, y) be the moving point. A quadratic graph is that of a parabola. The parabola is the locus of a point P which moves such that the distance of the point from focus and the directrix is equal. Here it is given that the focus is S(4, 3) and the directrix is y = k = 13. Draw PM perpendicular to y = k = 13 then, coordinates of M(x, 13).

By definition and the diagram,

PS = PM

Squaring both the sides,

PS^{2} = PM^{2}

(x - 4)^{2 }+ (y - 3)^{2}= (x - x)^{2 }+ (y - 13)^{2 }(using the distance formula between two points)

x^{2} - 8x + 16 + y^{2} - 6y + 9 = y^{2} - 26y + 169

x^{2} - 8x + 16 = -20y + 160

x^{2} - 8x + 16 = -20(y - 8)

(x - 4)^{2} = -20(y - 8), which is of the form (x - h)^{2} = -4a(y - k).

## What is the equation of the quadratic graph with a focus of (4, 3) and a directrix of y = 13?

**Summary:**

The equation of the quadratic graph with a focus of (4, 3) and a directrix of y = 13 is (x - 4)^{2} = -20(y - 8).

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